我有一个自定义 ContainerSegue
继承自 UIStoryboardSegue
。在 perform()
方法中,我需要访问执行当前 segue 的 UIView
。我怎样才能做到这一点?可能吗?
In other words I need to access the sender within
perform()
method.
最佳答案
segue
类通常不会跟踪 sender
,但是由于您已经定义了一个名为 的自定义
,你可以给它添加一个segue
类ContainerSeguesender
属性:
class ContainerSegue: UIStoryboardSegue {
// Add this property to hold the sender
var sender: AnyObject?
override func perform() {
if let button = sender as? UIButton, title = button.currentTitle {
print("button title is \(title)")
}
// Add remainder of perform code here
}
}
然后在 prepareForSegue
中设置:
override func prepareForSegue(segue: UIStoryboardSegue, sender: AnyObject?) {
if let containerSegue = segue as? ContainerSegue {
containerSegue.sender = sender
}
}
类似的,如果要访问destinationViewController
中的sender
,在destinationViewController中添加一个
并在 sender
属性prepareForSegue
中设置:
override func prepareForSegue(segue: UIStoryboardSegue, sender: AnyObject?) {
if let dvc = segue.destinationViewController as? MyDestinationVC {
dvc.sender = sender
}
}
关于ios - 如何从该 segue 或目标 View Controller 访问执行 segue 的源 View ?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38914846/